Chapter Three:
Stoichiometry: Calculations with
Chemical Formulas and Equations
Overview
Chemical Equations
 Patterns/Reactions
 Atomic/Molecular Weights
 Moles/Molar Mass
 Empirical/Molecular Formulas
 Quantitative Relationships
 Limiting Reactants/Theoretical Yields

Chemical Equations

chemical ‘sentences’
– reactants and products described by formulas or
symbols combined with “punctuation”
reactant
formulas
2 H2(g)
coefficients
product
formula
‘react to form’
+
O2(g)
2 H2O(l)
physical state
“atoms can be neither created nor destroyed”
– all equations must be ‘balanced’ with the same number
of atoms on both sides of the reaction arrow
unbalanced
H2O
2H
+

O2
&
H 2O 2

3O
2H & 2O
balanced
2H2O
4H
+
&

O2
4O
=
2H2O2
4H & 4O
H2O
O2
H2O2
unbalanced
balanced
one formula unit
two formula units
Examples
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4
H2O(l)
2
2
2

Na(s) + H2O(l)  NaOH(aq) +
2
2
H2(g)


HBr(aq)+ Ba(OH)2(aq)  H2O(l) +
BaBr2(aq)
Patterns of Chemical Reactivity

Because elements are grouped by chemical
properties, their reactions can also be
grouped:
– alkali metals and water
specific
2K(s) + 2H2O(l)  2KOH(aq) + H2(g)
general
2M(s) + 2H2O(l)  2MOH(aq) + H2(g)
– Combustion in air
specific
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
general
CxHy +
O2(g) 
CO2(g) +
hydrocarbon
H2O(l)
– Combination Reactions
specific
2Mg(s) + O2(g)  2MgO(s)
general
X
+
Y

XY
– Decomposition Reactions
specific
CaCO3(s)

CaO(s) + CO2(g)
general
XY

X
+
Y
Name the Reaction

PbCO3(s)  PbO(s)
+ CO2(g)

C(s) + O2(g) 

2NaN3(s)  2Na(s) + 3N2(g)

2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(l)
CO2(g)
decomposition
combination
decomposition
combustion
Atomic and Molecular Masses

Amu scale
–
–
–

defined by assigning the mass of 12C as 12 amu exactly
1 amu = 1.66054 x 10-24 g
1 g = 6.02214 x 1023 amu
Average Atomic Masses
–
12C
98.892% abundant 13C 1.1108% abundant
(0.98892)(12 amu) + (0.01108)(13.00335 amu) = 12.011
amu
atomic mass
Formula and Molecular Masses
– sum of all atomic masses in the formula of an ionic or
molecular compound
vitamin C
C6H8O6
6 x
12.0=
72.0 amu
8 x
1.0=
8.0 amu
6 x
16.0
= 96.0 amu
176.0 amu
formula mass of vitamin C
(often called molecular mass)
Percentage Composition

Calculate the percent mass that each type
of atom contributes to a molecule
– % X = (no. X atoms)(X amu) x 100
formula mass cmpd
– C6H8O6
% C = (6)(12.01amu) x 100 = 40.94% C
176.0 amu
% H = (8)(1.01amu) x 100 = 4.59% H
176.0 amu
% O = (6)(16.00 amu) x 100 = 54.55% O
176.0 amu
The Mole


We can measure masses in amu but how do we
relate that to mass in grams? We define a
quantity of atoms – a mole – which has the same
mass in grams as the mass of the element in amu.
So how many atoms does it take to make, say,
1.00 g of H?
1.0 g H x
1 atom H @ 6.0 x 1023 atoms of H
1.7 x 10-24 g H
a mole
12.0 g C x
1 atom C
@ 6.0 x 1023 atoms of C
2.0 x 10-23 g C
Avogadro’s Number

6.02214 x 1023 units/mole
–
–
–
–
No. of atoms per mole of an element
No. of molecules per mole of molecular cmpd.
No. of formula units per mole of ionic cmpd.
No. of cows per mole of cows
Memorize this number & what it means!

1 C atom = 12 amu
1 mole C atoms = 12 g

1 Mg atom = 24 amu
1 mole Mg atoms = 24 g

1 CO molecule = 28 amu
1 mole CO molecules = 28 g

1 NaCl fm. unit = 58 amu
1 mole NaCl fm.units = 58g
Molar Mass

From this information we can define something
called the molar mass (MM) of an atom (or
molecule or formula):
from the equality:
1 mole C = 12.0 g C
we define the molar mass of a substance
12.0 g C =
1 mole C
MM
or
conversion factor
Molar Mass
(Atomic Mass)
(Molecular Mass)
(Formula Mass)
Problems

Practice Ex. 3.9:
– How many mole in 508 g of NaHCO3?
Given:
MM = 84.02 g/mol NaHCO3 508 g NaHCO3
508 g NaHCO3 x 1 mole
=
84.02 g NaHCO3
6.05 mole NaHCO3
– How many formula units of NaHCO3?
Given:
6.02 x 1023 form. units/mole NaHCO3
6.05 mole NaHCO3 x 6.02 x 1023 fm. units = 3.6 x 1024 fm.
1 mole
units NaHCO3


Molar Mass converts between moles and
grams of a substance
Avogadro’s number converts between moles
of a substance and atoms (or molecules or
formula units) of that substance
These are very important conversion
factors, know & understand them!
Problems

How many moles of vitamin C are contained in 5.00 g of
vitamin C? C6H8O6 176.0 g/mol

17.5 mg of cocaine (C17H21NO4) per kg of body weight
is a lethal dose. How many moles is that? How many
molecules?

In 25 g of C12H30O2 THC (tetrahydrocannibinol) how
many moles are there? How many molecules are there?
How many C atoms are there?


How many moles of O are contained in 1.50
moles of C6H5NO3?
How many grams of nitrogen are contained
in 70.0 g of C6H5NO3? How many atoms?
Calculate the number of H
atoms in 50.0 mg of
acetominophen, C8H9O2N.
Determination Empirical Formulas

simplest ratio of atoms
– change g of each element to moles or
– assume 100 grams of substance & change
the % of each element to moles
– change the mole ratio of atoms to the
simplest ratio by dividing by the smallest
number of moles

Practice Ex. 3.12:
– 5.325 g methyl benzoate contains 3.758 g C, 0.316 g H,
1.251 g O. Determine empirical formula.
3.758 g C x
1 mole = 0.313 mol C
12.01 g
C0.313H0.313O0.0782
0.316 g H x
1.251 g O x
1 mole = 0.313 mol H
1.01 g
1 mole = 0.0782 mol O
16.00 g
C4H4O
Determination of Molecular Formulas

actual ratio of atoms
– determine the empirical formula
– divide the actual molar mass by the
empirical formula mass to get ‘n’
– multiply the mole ratio in the empirical
formula by ‘n’

Practice Ex. 3.13:
– Ethylene glycol is composed of 38.7% C, 9.7% H &
51.6% O by mass. Its true molar mass is 62.1 g/mol.
What are the empirical and molecular formulas?
38.7 g C x
9.7 g H x
51.6 g O x
1 mole
12.0 g
1 mole
1.01 g
1 mole
16.0 g
= 3.23 mole C
C3.23H9.60O3.22
= 9.60 mole H
empirical formula
= 3.22 mole O
CH3O
n=2
molecular formula
C2H6O2
Formulas from Combustion Data

Formulas determined from products of
combustion products
– Menthol is composed of C, H, and O. A 0.1005 g sample
of menthol is combusted, producing 0.2829 g of CO2 and
0.1159 g H2O. What is the empirical formula?
CxHyOz + O2
0.1005 g
 CO2 +
0.2829 g
H2O
0.1159 g
– Calculate moles CO2 & C; moles H2O & H
0.2829 g CO2 x 1 mol x 1 mol C = 0.00643 mol C
44.0 g
1 mol CO2
0.1159 g H2O x 1 mol x 2 mol H = 0.0129 mol H
18.0 g
1 mol H2O
total mass of C + H = 0.0902 g
mass of O = 0.1005 g - 0.0902 g = 0.0103 g O x 1 mol =
16.0 g
total mass of all C, H & O
6.44 x 10-4 mol O
0.00643 mol C
0.0129 mol H
6.44 x 10-4 mol O
C0.00643H0.0129O0.000644
C10H20O
(empirical formula mass 156 g/mol)
If the MM is 156 g/mol, what is the molecular formula?
n=1 therefore molecular formula is C10H20O
Quantitative Stoichiometry

Determination of quantities from balanced
chemical reaction equations
– mole ratios from balanced chemical equation
convert between species
– if quantities are given for more than one reactant,
the limiting reactant must be determined
– Given the following balanced equation:
1Mg(OH)2 + 2HCl  1MgCl2 + 2H2O
– Calculate the number of moles of HCl required to
react completely with 0.42 mol of Mg(OH)2
0.42 mol Mg(OH)2 x
2 mol HCl
=
1 mol Mg(OH)2
0.84 mol HCl
The mole ratio comes from the balanced chemical
equation
– How many grams of MgCl2 can be produced?
0.42 mol Mg(OH)2 x 1 mol MgCl2 x
1 mol Mg(OH)2
Theoretical Yield -- maximum
amount that can be produced
95.3 g MgCl2
1 mol
= 40.0 g MgCl2
General Sequence of Conversion:
grams of
reactant
MM reactant
moles of
reactant
mole
ratio
grams of
product
MM product
moles of
product
Practice Ex. 3.14:

How many grams of O2 can be prepared
from 4.50 g of KClO3?
2KClO3
 2KCl + 3O2
4.50 g KClO3 x 1 mol x 3 mol O2 x 32.0 g O2 = 1.76 g O2
122.6 g 2 mol KClO3 1 mol
Limiting Reactant

given a non-stoichiometric amount of both
reactants, you will have to determine which is the
limiting reagent or reactant

example: you have 10 bicycle frames and 16 bicycle
wheels and you need to put them together to produce as
many bicycles as possible, how many bicycles can be
produced, what is the limiting “reagent”, and how much
excess “reagent” do you have left over?
— Balanced
‘Equation’
1 (mole) frame + 2 (moles) wheels 1 (mole) bicycles
[10 (moles) frames]
[16 (moles) wheels]
[8(moles) bicycles]
– Limiting Reactant -- will produce the least amount of
product
10 mol frames x 1 mol bicycles = 10 bicycles
1 mol frames
limiting reactant
16 mol wheels x 1 mol bicycles = 8 bicycles
2 mol wheels
Practice Ex. 3.16:

A mixture of 1.5 mol of Al and 3.0 mol of Cl2 react.
What is limiting & how many moles of AlCl3 are
formed?
2Al(s) +
3Cl2(g)  2AlCl3(s)
1.5 mol
3.0 mol
1.5 mol Al x 2 mol AlCl3
2 mol Al
3.0 mol Cl2 x 2 mol AlCl3
3 mol Cl2
1.5 mol
= 1.5 mol AlCl3
= 2.0 mol AlCl3